Difference between revisions of "2001 AMC 10 Problems/Problem 2"
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<math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>. | <math> x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1</math>. | ||
Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>. | Therefore, <math> \boxed{\textbf{(C) }0 < x\le 2} </math>. | ||
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+ | ==Video Solution by Daily Dose of Math== | ||
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+ | https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS | ||
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+ | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:09, 15 July 2024
Problem
A number is more than the product of its reciprocal and its additive inverse. In which interval does the number lie?
Solution
We can write our equation as . Therefore, .
Video Solution by Daily Dose of Math
https://youtu.be/6wBIYhmCfo8?si=tRO5zDw7syrXCPsS
~Thesmartgreekmathdude
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.